poj1149 最大流 PIGS

最新推荐文章于 2019-12-18 20:07:00 发布

原创最新推荐文章于 2019-12-18 20:07:00 发布 · 569 阅读

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本文探讨了如何运用网络流算法解决一个养猪场的销售问题，通过构建网络模型，实现最大化的销售数量。文章详细介绍了算法的构建过程，包括顾客与猪圈之间的连接方式、边权设定及最终求解最大销售量的方法。

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题意：M个猪圈，N个顾客，每个顾客有一些的猪圈的钥匙，只能购买这些有钥匙的猪圈里的猪，而且要买一定数量的猪，每个猪圈有已知数量的猪，
但是猪圈可以重新打开，将猪的个数，重新分配，以达到卖出的猪的数量最多。

思路：刚学网络流，表示很菜很菜很菜~~
①构造网络，将顾客看成源点和汇点以外的结点，并设另外两个节点：源点和汇点。
②源点和每个猪圈的第一个顾客连边，边的权是开始时候猪圈中猪的数量。
③ 若源点和某个节点之间有重边，则将权合并
④顾客j紧跟顾客i之后打开某个猪圈，则<i.j>的权是正无穷。
⑤每个顾客和会点之间连边，边的权值是顾客所希望购买的猪的数量。
例如：样例中的就可以建立如图：

其中inf是正无穷~~

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

Sample Output

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
#define maxn 1005
#define INF 0x3f3f3f3f
int cap[maxn][maxn],flow[maxn][maxn];
int a[maxn],p[maxn];
int n,m;
int EK()
{
    int s=0,t=n+1;
    queue<int>q;
    int f=0;
    for(;;)
    {
        memset(a,0,sizeof(a));
        a[s]=INF;
        q.push(s);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int v = 0; v <=t; v++)
                if(!a[v] && cap[u][v] > flow[u][v])
                {
                    p[v] = u;
                     q.push(v);
                    a[v] = min(a[u], cap[u][v]-flow[u][v]);

                }
        }
        if(a[t]==0)
            break;
        for(int u=t; u!=s; u=p[u])
        {
            flow[p[u]][u]+=a[t];
            flow[u][p[u]]-=a[t];
        }
        f+=a[t];
    }
    return f;
}
int main()
{
    int pig[maxn]={0},vist[maxn]={0};
    int k,key;
    memset(cap,0,sizeof(cap));
    scanf("%d %d",&m,&n);
    for(int i=1;i<=m;i++)
        scanf("%d",&pig[i]);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&k);
        for(int j=1;j<=k;j++)
            {
                scanf("%d",&key);
        if(!vist[key])
            cap[0][i]+=pig[key];
            else

                cap[vist[key]][i]=INF;
            vist[key]=i;
        }
        scanf("%d",&key);
        cap[i][n+1]=key;
    }
    printf("%d\n",EK());
    return 0;
}